Projective plane as quotient space of a sphere

First of all, I'm not sure if this is the right forum, but none of the forums mention topology in their description. But anyway, I'm taking a topology class, and the professor mentioned that the projective plane is obtained by identifying antipodal points on the sphere, ie, points diametrically opposite to each other. To get a better sense of what the space was like, he said it could also be obtained by considering just one half of the sphere and identifying opposite points on the boundary (equator). This seems reasonable, but I can't seem to come up with a rigorous proof that the two quotient spaces are homeomorphic. Can anyone help me out?

Well P can be defined as the quotient group of the 2-sphere, S, and the equivalence relation which identifies antipodal points. P=S/~. The antipodal mapping is just f(x)=-x. So f: S--->P=S/~ given by x l----> -x is the desired homeomorphism? You could also choose the homeomorphism [x,y] l----> x/y which maps P to S, where [x,y] is the equivalence class determined by (x,y) such that if a=cx, b=cy, then a,b are elements of [x,y].

Topologists often don't bother constructing direct homeomorphisms if a couple of pictures will do the same.

Take the 2-sphere and cut it along the equator (make an "open northern hemisphere) (we're carrying out the antipodal identifiction here) take the northern hemisphere twist it 180 degrees and turn upside down. Plunk it down into the southern hemisphere. You've now identifed all the points not on the equator to all their antipodal points. There's only one step left, identifying the antipodal points on the equator, that's why their the same.

Notice this is the same as identifying antipodal points on the boundary of a disc.

a continuous map from the sphere with antipodal points identified is a continujos map from the sphere which sends antip[odal points to the samer point.

so map each point of the sphere to the point in the upper half sphere which lies on the same line through the origin as the original point.

then define a map back. send each point of the upper half sphere to itself in the full sphere, then compose with the identification map from the whole sphere to projective space.

see if thpose maps are inverse to each other, if so you have your homeomorphism. and never let yourself be put off from your question by an answer like "topologists do not stoop to actually giving a homeomorphism as they are too cool to do so."

it may be true, but if it does not apply to you, just ask for an honest answer to your question.

Here's a more "visual" presentation. The "points" of projective space are the parallelism classes of lines in 3-space. Each class can be repesented by a single line through one selected point (the origin), with direction on the lines not defined. Then since length is not delimited we can choose the representatives all to have the same length for convenience, and then they become the diameters of a 2-sphere with center at the origin. Each diameter defines two antipodal points on the sphere, and I reiterate that each diameter is a rep of one point in the projective pane, so each pair of antipodal points also represents one point.

"topologists do not stoop to actually giving a homeomorphism as they are too cool to do so."

I was only pointing out the custom that I have observed, but passed no judgement on it. Many times constructing an explicit form of a homeomorphism can be daunting, when other methods are preferrable. I certainly agree that anyone who wishes to, should do so. But only gave my advice to point out what is commonly done, and how the homeomorphism can be understood without an explicit form.

All of your replies were helpful. I wanted to see how a rigorous proof would look so that I'm more comfortable relying on intuition in other cases. But homology's construction made it easier to see why a homeomorphism should exist in this case. By the way, can anyone answer my other question about where topology questions should be posted?